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Topic: Order embedding
Replies: 8   Last Post: Sep 16, 2013 10:01 PM

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David C. Ullrich

Posts: 3,224
Registered: 12/13/04
Re: Order embedding
Posted: Sep 16, 2013 11:32 AM
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On Mon, 16 Sep 2013 14:24:06 +0300, Victor Porton <porton@narod.ru>
wrote:

>William Elliot wrote:
>

>> Let X,Y be (partially) ordered sets. Are these definitions correct?
>>
>> f:X -> Y is order preserving when
>> for all x,y, (x <= y implies f(x) <= f(y).

>
>Yes.
>

>> f:X -> Y is an order embedding when
>> for all x,y, (x <= y iff f(x) <= f(y)).

>
>Yes.


I think no. Surely an embedding is required to be injective.

>
>> f:X -> Y is an order isomorphism when f is surjective
>> and for all x,y, (x <= y iff f(x) <= f(y)).

>
>Yes.


No. An isomorphism must be bijective.

>> The following are immediate consequences.
>>
>> Order embedding maps and order isomorphisms are injections.

>
>Yes.
>

>> If f:X -> Y is an order embedding,
>> then f:X -> f(X) is an order isomorphism.

>
>Yes.
>

>> Furthermore the composition of two order preserving, order
>> embedding or order isomorphic maps is again resp., order

>
>> preserving, order embedding or order isomorphic.
>
>Yes.
>

>> Finally, the inverse of an order isomorphism is an order isomorphism.
>
>Yes.
>

>> That all is the basics of order maps, is it not?
>> Or is the more to be included?

>
>Probably all.
>
>There are also Galois connections.





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